"""
Write an iterator that iterates through a run-length encoded sequence.

The iterator is initialized by RLEIterator(int[] A), where A is a run-length encoding of some sequence.  More specifically, for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is repeated in the sequence.

The iterator supports one function: next(int n), which exhausts the next n elements (n >= 1) and returns the last element exhausted in this way.  If there is no element left to exhaust, next returns -1 instead.

For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of the sequence [8,8,8,5,5].  This is because the sequence can be read as "three eights, zero nines, two fives".

 

Example 1:

Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
Output: [null,8,8,5,-1]
Explanation: 
RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
This maps to the sequence [8,8,8,5,5].
RLEIterator.next is then called 4 times:

.next(2) exhausts 2 terms of the sequence, returning 8.  The remaining sequence is now [8, 5, 5].

.next(1) exhausts 1 term of the sequence, returning 8.  The remaining sequence is now [5, 5].

.next(1) exhausts 1 term of the sequence, returning 5.  The remaining sequence is now [5].

.next(2) exhausts 2 terms, returning -1.  This is because the first term exhausted was 5,
but the second term did not exist.  Since the last term exhausted does not exist, we return -1.

Note:

0 <= A.length <= 1000
A.length is an even integer.
0 <= A[i] <= 10^9
There are at most 1000 calls to RLEIterator.next(int n) per test case.
Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.

实现一个RLE。
什么是RLE呢：
111222333
会被压缩为
3个13个23个3这样的。

题目是给出的压缩后的，next会返回第x个数，不够的话返回-1。

直接迭代然后判断数量即可。

测试地址：
https://leetcode.com/contest/weekly-contest-101/problems/rle-iterator/

contest，运行了76ms。

"""
class RLEIterator(object):

    def __init__(self, A):
        """
        :type A: List[int]
        """
        self.a = A

    def next(self, n):
        """
        :type n: int
        :rtype: int
        """
        for i in range(0, len(self.a), 2):
            if self.a[i] > 0:
                if self.a[i] - n >= 0:
                    self.a[i] -= n
                    return self.a[i+1]
                else:
                    t = self.a[i]
                    self.a[i] -= n
                    n -= t
        return -1


# Your RLEIterator object will be instantiated and called as such:
# obj = RLEIterator(A)
# param_1 = obj.next(n)
